When $\mathrm{SO}_{2}$ is passed through acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{0}$, the process…

When $\mathrm{SO}_{2}$ is passed through acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{0}$, the process that takes place is
  1. the solution turns blue
  2. the solution is decolourised
  3. $\mathrm{SO}_{2}$ is reduced
  4. green $\mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}$ is formed

Solution

When $\mathrm{SO}_{2}$ is passed through acidified $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}$ solution. $\mathrm{SO}_{2}$ gas turns acidified potassium dichromate solution from orange to green reduced chromium $+4$ to $+3$. $\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}+2 \mathrm{H}_{2} \mathrm{SO}_{4}+3 \mathrm{SO}_{2} \rightarrow 2 \mathrm{Cr}_{2}\left(\mathrm{SO}_{4}\right)_{3}+\mathrm{K}_{2} \mathrm{SO}_{4}+\mathrm{H}_{2} \mathrm{O}$

Asked in: MHT CET 2020 (14 Oct Shift 1)

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